A technique to solve Bending Stress and Shear stress in an Asymmetric I-Section.For Civil and Mechanical Engineering students.

1. Determination of Bending Stress
and Shear Stress in a Beam
Subjected to Pure Bending
By
Prof.Abhinav
abhinav.t@alliance.edu.in
Alliance College of Engineering and Design

2. 240 mm
180 mm
G1
G2
G3
Unsymmetrical I Section
x x
10mm
300mm
10mm
10mm
1
2
3
1.Determination of Centroid of the individual section 1,2
and 3.(taking reference line xx) as shown in the Fig.1
g1
g2
g3
y1=10+300+10/2=315 mm
y2=10+300/2=160 mm
y3=10/2=5 mm
2.Determination of Centroid of complete section (taking
reference line xx) as shown in the Fig.1
An unsymmetrical I section is shown in the Fig.1, is subjected to
BM of 15kN-m.The top flange is in compression .Draw the bending stress
and shear stress diagram across the section marking the values at the
salient point.
Y= (a1×y1+a2×y2+a3×y3)/(a1+a2+a3)=172.92 mm ,w.r.t bottom layer
a1=240×10=2400 mm2
a2= 10×300=3000 mm2
a3= 10×180=1800 mm2
G
3.Determination of the distance of the extreme fiber from the neutral axis
NA
172.92 mm
From the axis xx
ymaxT =172.92 mm
ymaxC = (10+300+10 )-YmaxT
=147.08 mm
4. Calculation of Moment of Inertia using parallel axis theorem of I beam and is give by as follows
I= [b1d1
3/12+a1h1
2]+ [b2d2
3/12+a2h2
2]+ [b3d3
3/12+a3h3
2]
Ix =Ixx+ ah2
Where h is the distance between centroidal axis of section to the axis of Interest(xx)
Centroidal axis of section
147.08 mm
b
d
y
Content belongs to Prof. Abhinav, Department of Mechanical Engineering
A A
B BC C
D DE E
F F

4. Where h is the distance between centroidal axis of the section to the centroidal axis of the individual section
I= [b1d1
3/12+a1h1
2]+ [b2d2
3/12+a2h2
2]+ [b3d3
3/12+a3h3
2]
h1=y1
y2
y3
y1- y = 315-172.92=142.08 mm
h2= y - y2
y
h3= y – y3
=172.92-160=12.92 mm
=172.92- 5= 167.7 mm
I= 48.47×106 + 23×106 +50.77×106
=122.24×106
5. Determination of the magnitude of maximum stress in compression fmaxC and in Tension fmaxT
M
I
fmaxC
YmaxC
=
We know from Pure Bending equation or Bernoulli's equation for the determination of Maximum Bending Stress
fmaxC × YmaxC
M
I
=
(15×106 ×147.08) /122.24×106
= 18.05 N/mm2fmaxC =
M
I
fmaxT
YmaxT
=
(15×106 ×172.92) /122.24×106
= 21.22 N/mm2fmaxT =
fmaxT × YmaxT
M
I
=
fmaxC =
fmaxT
Content belongs to Prof. Abhinav, Department of Mechanical Engineering

5. 240 mm
180 mm
G1
G2
G3
Unsymmetrical I Section
x x
10mm
300mm
10mm
10mm
1
2
3
g1
g2
g3
An unsymmetrical I section is shown in the Fig.1, is subjected to
BM of 15kN-m.The top flange is in compression .Draw the bending stress and
shear stress diagram across the section marking the values at the salient
point.
G
NA
172.92 mm
From the axis xx
Centroidal axis of section
147.08 mm
b
d
y
fmaxC =
fmaxT
M f’c
I y
=
y
y
6.Determination of bending stress at the top section
Determination of bending stress at the bottom section
y= 147.08-10 =137.08 mm
M f’t
I y
=
y= 172.92-10 =162.92 mm
f’c = 16.82 N/mm2
f’c
f’t = 19.99 N/mm2
f’t
f’c = 16.82 N/mm2
f’t = 19.99 N/mm2
fmaxC =18.05 N/mm2
Content belongs to Prof. Abhinav, Department of Mechanical Engineering
A A
B BC C
D DE E
F F fmaxT =21.22 N/mm2

6. q at AA= 0 N/mm2
q at BB =
7.Determination of Shear Stress magnitude and plot when the magnitude of shear stress is given say 50KN
Shear stress will be always zero t the extreme planes
Q 2. Cross section of the beam is shown in the Fig. The section is subjected to a shear force of 50KN.
Draw the shear stress distribution diagram given y =166.51 mm from bottom and I about NA=2.849×108 mm4
Fay
Ib Where
F is the shear force
a is the area
Y is the distance from the centroid of the complete section to the centroid of the individual section
GNA
y =166.51 mm
50 mm
50 mm
200 mm
(300-166.51)=133.49 mm
A A
B BC C
D DE E
F F
1
2
3
X X
y =(133.49-50/2)= 108.49 mm
[50×103×(200×50) × 108.49]/2.89 ×108
= 0.95 N/mm2
130 mm
200 mm
q at CC= 0.95 ×(200/50)= 3.81 N/mm2
50 mm
y=108.49 mm
q at NA = (50 ×103 × 1.259 ×106)/2.849 × 108 ×50= 4.42 N/mm2
Where summation of ay of section 1 and 2 i.e.(There are two sections)
= [(200×50) ×108.49]+[(50×83.49) ×(83.49/2)]= 1.259 ×106 mm3
(250-166.51)= 83.49 mm
q at EE = (50 ×103 × 0.919 ×106) / (2.849 × 108 ×130)= 1.42 N/mm2
There is only one section (130 × 50) ×(166.51-50/2)= 0.919 ×106 N/mm3
q at DD= 1.24 ×130/50 =3.22 N/mm2
q at FF= 0 N/mm2
Parabolic variation
Parabolic variation